Domain of a Composition Definition Given the function f and g, the composition of f with g is a function defined as (f g)() f(g()). The domain of f g is the set of all real numbers in the domain of g such that g() is in the domain of f. 2 Eplanation We have seen that determining the domain of a composition can be a bit difficult, so let s try to break down the above definition to get a better idea of what we are actually looking for. First, it is clear we need to find the domain of g: The domain of f g is the set of all real numbers in the domain of g.... This step shouldn t be too difficult, just remember that 99% of the time you will be looking for which -values give you a zero in a denominator, or make you take the square-root (or any even-root, i.e. 4th-root, 6th-root, etc.) of a negative number (these -values will not be in the domain). The latter portion of the definition is where the difficulty generally occurs:...g() is in the domain of f. Let us begin by calculating the domain of f (again this should not be too difficult), but instead of writing the domain in interval notation (as we usually do) let s use inequalities, i.e. <,, >,,. So now we have conditions on which -values are in the domain of f, but when we look back at the definition see that we are not interested in the -values in the domain of f, but rather the g()-values in the domain of f. Thus, substitute g() in for inside of the inequalities you found for the domain of f. Once you have made this substitution, solve for. These will now be the -values for which g() is in the domain of f (which is what we want). After doing this we know the -values in the domain of g, and we know which -values keep g() in the domain of f (it may be easier at this point to write these -values in interval notation). Hence, to find the domain of f g, we simply need to intersect these two intervals (i.e. find the -values which are in both of our intervals). Eamples As always it s easier to understand a new concept if you see a few eamples, so we now will work through a few problems using the above eplanation. Eample Let f() +2 and g(). Find (f g)() and determine its domain.
(f g)() f(g()) f ( ) + 2 +2 + 2 2 2. First we will determine the domain of g. Since g() 4 we see that if 0, then we will have a 0 in the denominator, and hence 0 is not in the domain of g. Therefore, Domain of g (, 0) (0, ) (i.e. 0). Now we want to find the domain of f. Since f() +2 in the denominator, so 2 is not in the domain of f. Thus, Domain of f (, 2) ( 2, ) we see that if 2 we will have a 0 (i.e. 2). Remember that we don t want to know which s are in the domain of f, but rather which g() s are in the domain of f. Hence, we substitute g() in for to find, Solving this for we see g() 2. 2 2. This means that for 2 (in interval notation: (, 2) (2, )) we will have that g() is in the domain of f. Finally, we need to find the intersection of our two intervals (i.e. find all numbers that are in both intervals): Domain of f g (Domain of g) (-values where g() is in the domain of f) ((, 0) (0, )) ((, 2) (2, )) (, 0) (0, 2) (2, ) { 0, 2}. Eample 2 Let f() 2 + 8 and g(). Find (f g)() and determine its domain. (f g)() f(g()) f( ) ( ) 2 + 8 + 8 +. First we will determine the domain of g. Since g() we see that if < we will take the square root of a negative number, so these values are not in the domain. Therefore, Domain of g [, ) (i.e. ). Now we want to find the domain of f. Since f() 2 + 8, we see that there are no square roots or denominators, so the domain of f is the set of all real numbers, Domain of f (, ) (i.e. < < ). Again, we are only interested in when g() is in the domain of f, but this time things are very easy. We know that every real number is in the domain of f, so as long as g() is a real number it will be in the domain of f. We know the only times that g() fails to be a real number are when is not in the domain of g. Hence, g() will be in the domain of f when is in the domain of g, i.e. in the 2
interval [, ). Thus, we look at the intersection of our intervals to determine the domain of f g: Domain of f g (Domain of g) (-values where g() is in the domain of f) ([, )) ([, )) [, ) { }. Now we will work though two eamples which are more difficult. Eample Let f() and g() 2. Find (f g)() and determine its domain. (f g)() f(g()) f( 2 ) 2. First we will determine the domain of g. Clearly the domain of g will be all real numbers (there are no denominators or square roots), so Domain of g (, ) (i.e. < < ). Now we will determine the domain of f. Since f() we know that if > we will have a square-root of a negative number, so these values will not be in the domain. Hence, Domain of f (, ] (i.e. ). Again, we want to know when for which -values g() will be in the domain of f, so we make a substitution: 2 g(). This is where things get tricky. If we were to solve for as we usually do, we would most likely take the square-root of both sides to find ±. If we wrote this in interval notation, we would have (, ] (since we would require both and at the same time), but there are obviously numbers inside that interval which violate the condition 2. For instance, 2 is inside the interval (, ], but ( 2) 2 4 which is definitely not less than or equal to! So we need to try a different approach. Instead of just jumping in and solving for, let us take a moment and think about what the inequality 2 is telling us. The inequality is saying that if we multiply a number by itself, the resulting product is less than or equal to. If we pick a number larger than and multiply it by itself, we will alway get a number that is larger that. For instance,.00000000 > and (.00000000) 2.00000000200000000 >. Also, we should note that 2 is an even function, which means ( ) 2 2. Hence, if we pick any negative number and square it, it would be the same as squaring the positive version of that number. Hence, if we take any number less than and square it, the result will again be greater than. Thus, we may rule our all values with > and <. It is clear that both and satisfy our condition 2. You should also note that the square of any number between 0 and is actually smaller than your original number! For instance, (0.2) 2 0.062. So any number between 0 and will satisfy our condition. Again using the fact that 2 is an even function, it follows that any number between and 0 will also satisfy our condition.
Therefore, putting all of this together, the only -values which satisfy 2 are the -values such that (i.e. inside the interval [.]). Hence, the -values where g() is in the domain of f are in the interval [, ]. So, we find Domain of f g (Domain of g) (-values where g() is in the domain of f) ((, )) ([, ]) [, ] { }. Eample 4 Let f() + and g() 2. Find (f g)() and determine its domain. (f g)() f(g()) f ( ) + ( 2) 2 + 2 2. First we will determine the domain of g. Since g() 2 we know that if 2 we will have a zero in the denominator, so 2 is not in the domain of g. Thus, Domain of g (, 2) (2, ) (i.e. 2). Now we will determine the domain of f. Since f() +, we know that if < we will have the square-root of a negative number, and hence these values will not be in the domain. Hence, Domain of f [, ) (i.e. ). We are only interested in when g() is in the domain of f, so we make our usual substitution: g(). 2 Again, we run into a slight problem here. We want to solve for, but we need to remember that if you have an inequality and multiply both sides by a negative number, you must change the direction of the inequality. Note that if < 2 then 2 will be negative! So if < 2 and we multiply both sides by 2 we must remember to change the direction of our inequality! Thus, we need to break our problem up into two cases: when > 2 and when < 2 (note, we do not need to consider 2, since this would give us a zero in the denominator). Let us first suppose that > 2, then we know that 2 0, so we can multiply both sides of our inequality without needing to do anything special: 2 ( 2) 2 ( 2) + 6. To keep going, we will need to multiply both sides by, and since this is negative, we must change the direction of our inequality!. At the start we required > 2 and we just found that we need for g() to be in the domain of f, so putting these inequalities together we find that we need > 2 for g() to be in the domain 4
of f (i.e. we need inside the interval (2, )). (Note, we also could have done this case mentally. If > 2 then we know 2 > 0 and so 2 > 0 > ; hence so every > 2 satisfies our inequality, which is what we found above!) Now we need to consider our other case, < 2. When < 2 we stated above that we must change the direction of our inequality if we multiply by 2. So we see: 2 ( 2) 2 ( 2) + 6. Again, we will multiply by, so we need to change the direction of our inequality!. At the start we required < 2, and we just found that we need for g() to be in the domain of f. So, if we put these inequalities together, we find that we need for g() to be in the domain of f (i.e. we need inside the interval (, ]). Putting our two cases together, we have found that if > 2 or, then g() will be in the domain of f; which in interval notation can be written (, ] (2, ). So finally we can determine the domain of f g: Domain of f g (Domain of g) (-values where g() is in the domain of f) (( ((, 2) (2, )), ] ) (2, ) (, ] (2, ) {, > 2}.